Hydrogen peroxide solution `(20 mL)` reacts quantitatively with a solution of `KMnO_(4) (20 mL)` acidified with dilute of `H_(2)SO_(4)`. The same volume of the `KMnO_(4)` solution is just decolourised by `10 mL` of `MnSO_(4)` in neutral medium simultaneously forming a dark brown precipitate of hydrated `MnO_2`. The brown precipitate is dissolved in `10 mL` of `0.2 M` sodium oxalate under boiling condition in the presence of dilute `H_(2)SO_(4)`. Write the balanced equations involved in the reactions and calculate the molarity of `H_(2)O_(2)`.

Explanation

m.e of `H_(2)O_(2)` in 20 mL solution
= m.e of 20 mL of `KMnO_(4)`
= m.e of 20 mL of `MnsO_(4)`
= m.e of `MnO_(2)`
= m.e of `Na_(2)C_(2)O_(4)`
` = 0.4 xx 10 ` (normality = molarity `xx` change in ON)
= 4
Eq. of `H_(2)O_(2)` in 20 mL solution
` = 4/1000`
Mole of `H_(2)O_(2)//20 mL = 4/1000 xx1/2 `
Molarity (mole.L) = `4/2000 xx 1000/20`
`0.1 M `
The reactions involved are ,
`5H_(2)O_(2)+2KMnO_(4)+3H_(2)SO_(4)=2MnSO_(4)+K_(2)SO_(4)+8H_(2)O+5O_(2)`
`2KMnO_(4) +3MnSO_(4) +2H_(2)O = 5MNO_(2) +2H_(2)O `
`MnO_(2) +Na_(2)C_(2)O_(4) +2H_(2)SO_(4) = MnSO_(4) +Na_(2)SO_(4) +CO_(2)+2H_(2)O `